An Insurance agent claims that the average age of policy holders who insure through him is less than the average of all other agents, which is 30. A random sample of 100 policy holders who had insured through the agent gave the following age distribution:
Test the claim that the average age is less than the figure for all the other agents at a 5% level of significance.
Answer (2)
Calculate the sample mean age from the given distribution: x ˉ = 32.2 .
Calculate the sample standard deviation: s = 9.4 .
Perform a one-tailed t-test with null hypothesis H 0 : μ = 30 and alternative hypothesis H 1 : μ < 30 .
Calculate the t-statistic: t = 2.3404 . Since -1.660"> t > − 1.660 (critical value), we fail to reject the null hypothesis. Fail to reject the null hypothesis.
Explanation
Analyze the problem First, let's analyze the problem. We are given the age distribution of 100 policyholders insured through an agent and we want to test the claim that the average age of these policyholders is less than 30 years, which is the average age of policyholders of all other agents. We will perform a hypothesis test with a significance level of 5%.
Calculate the sample mean Next, we need to calculate the sample mean and standard deviation from the given age distribution. The age groups are 16-20, 21-25, 26-30, 31-35, 36-40, 41-45, and 46-50. We will use the midpoints of these intervals as the representative ages for each group: 18, 23, 28, 33, 38, 43, and 48. The number of persons in each group is 12, 16, 18, 20, 10, 12, and 12, respectively.
To calculate the sample mean x ˉ , we use the formula: x ˉ = ∑ i = 1 n f i ∑ i = 1 n f i x i where x i is the midpoint of each age group and f i is the number of persons in that group.
Calculate the sample mean value Using the provided data, we calculate the sample mean: x ˉ = 100 ( 18 × 12 ) + ( 23 × 16 ) + ( 28 × 18 ) + ( 33 × 20 ) + ( 38 × 10 ) + ( 43 × 12 ) + ( 48 × 12 ) x ˉ = 100 216 + 368 + 504 + 660 + 380 + 516 + 576 = 100 3220 = 32.2 So, the sample mean age is 32.2 years.
Calculate the sample standard deviation Now, we calculate the sample standard deviation s using the formula: s = ∑ i = 1 n f i − 1 ∑ i = 1 n f i ( x i − x ˉ ) 2 However, since we will be using a t-test and the sample size is large (n=100), we can approximate the sample standard deviation using the formula: s = n − 1 ∑ i = 1 n f i ( x i − x ˉ ) 2
Using the calculated mean of 32.2, we find the standard deviation to be approximately 9.4.
State the null and alternative hypotheses We set up the null and alternative hypotheses:
H 0 : μ = 30 (The average age of the agent's policyholders is equal to 30 years) H 1 : μ < 30 (The average age of the agent's policyholders is less than 30 years)
We will use a one-tailed t-test because the alternative hypothesis is directional (less than).
Calculate the t-statistic Now, we calculate the t-statistic using the formula: t = s / n x ˉ − μ 0 where x ˉ = 32.2 , μ 0 = 30 , s = 9.4 , and n = 100 .
t = 9.4/ 100 32.2 − 30 = 0.94 2.2 ≈ 2.3404
Determine the degrees of freedom and critical value The degrees of freedom are df = n − 1 = 100 − 1 = 99 .
For a one-tailed t-test with df = 99 and α = 0.05 , the critical value is approximately -1.660.
Compare the t-statistic with the critical value We compare the calculated t-statistic (2.3404) with the critical value (-1.660). Since 2.3404 > -1.660, we fail to reject the null hypothesis. Also, the p-value is 0.9894, which is greater than 0.05, so we fail to reject the null hypothesis.
State the conclusion Conclusion: There is not enough evidence to support the claim that the average age of the agent's policyholders is less than 30 years at a 5% significance level.
Examples
In insurance, understanding the age distribution of policyholders is crucial for risk assessment and premium calculation. For instance, if an insurance agent claims to attract younger clients, this could imply lower risk and potentially justify different premium rates. Hypothesis testing, as demonstrated here, helps verify such claims by comparing sample data against established averages, ensuring fair and accurate insurance practices.
The insurance agent's claim that the average age of his policyholders is less than 30 years is not supported by the evidence from the sample. The t-test shows a t-statistic of approximately 2.3404, which does not exceed the critical value of -1.660, so we fail to reject the null hypothesis. Therefore, there is insufficient evidence to validate the agent's claim at a 5% significance level.
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